$ B = \left[\begin{array}{rr}4 & 3 \\ 3 & 1\end{array}\right]$ $ C = \left[\begin{array}{rr}-1 & 5 \\ -2 & 1\end{array}\right]$ What is $ B C$ ?
Explanation: Because $ B$ has dimensions $(2\times2)$ and $ C$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ B C = \left[\begin{array}{rr}{4} & {3} \\ {3} & {1}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{5} \\ {-2} & \color{#DF0030}{1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{-1}+{3}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-1}+{3}\cdot{-2} & ? \\ {3}\cdot{-1}+{1}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-1}+{3}\cdot{-2} & {4}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{1} \\ {3}\cdot{-1}+{1}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{-1}+{3}\cdot{-2} & {4}\cdot\color{#DF0030}{5}+{3}\cdot\color{#DF0030}{1} \\ {3}\cdot{-1}+{1}\cdot{-2} & {3}\cdot\color{#DF0030}{5}+{1}\cdot\color{#DF0030}{1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-10 & 23 \\ -5 & 16\end{array}\right] $